Answer
Both $(A_1) $ and $(A_2)$ fails
Work Step by Step
$(A_1)$ Let
$f(x)=a_0+a_1 x+a_2 x^2 \;\;\; \in S$
$g(x)=b_0+b_1 x+b_2 x^2 \;\;\; \in S$
By definition of $S$
$\Rightarrow a_0+a_1+a_2=1$______(1)
$\Rightarrow b_0+b_1+b_2=1$______(2)
$f(x)+g(x)=(a_0+b_0)+(a_1+b_1) x+(a_2+b_2) x^2$
Let $f(x)+g(x)=c_0+c_1 x+c_2 x^2$
Where $c_i=a_i+b_i$
Consider $c_0+c_1+c_2=(a_0+b_0)+(a_1+b_1)+(a_2+b_2)$
Because all $a_i$ and $b_i$ are real numbers
$c_0+c_1+c_2=(a_0+a_1+a_2) + (b_0+b_1+b_2)$
From (1) and (2)
$c_0+c_1+c_2=1+1=2$
So by definition of $S$ , $f(x)+g(x)$ is not in $S$
$\Rightarrow S$ is not closed with respect to addition.
$(A_2)$ Because we may choose scalars from set of real numbers
In particular , Let $\lambda =5$ be a scalar
and $f(x)=1-x+x^2 \:\in S$
$5.f(x)=5.(1-x+x^2)=5-5x+5x^2$
But sum of coefficients of $5.f(x)$ is 5$\neq 1$
So $5.f(x)$ is not in $S$
$\Rightarrow S$ is not closed with respect to scalar multiplication
